The elastic behaviour of a material for linear stress and linear strain is shown in the figure. The energy density for a linear strain of 5×10⁻⁴ is kJ m⁻³. Assume that the material is elastic up to the linear strain of 5×10⁻⁴.

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Accepted Answer

25. Idea: the graph is a straight line through the origin, so its slope gives the compliance and its reciprocal gives Y. From the plotted points, a stress of 20 N m⁻² produces a strain of 1×10⁻¹⁰, so Y=(20)/(1×10⁻¹⁰)=2×10¹¹ N m⁻². Within the elastic region the energy stored per unit volume is u=1/2Yε². u=1/2×2×10¹¹×(5×10⁻⁴)²=1/2×2×10¹¹×2.5×10⁻⁷. u=2.5×10⁴ J m⁻³=25 kJ m⁻³, so the answer is 25.

The elastic behaviour of a material for linear stress and linear strain is shown in the figure. The energy density for a linear strain of 5×10⁻⁴ is ______ kJ m⁻³. Assume that the material is elastic up to the linear strain of 5×10⁻⁴.

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